On the Nonexistence of Odd Perfect Numbers

ON THE NONEXISTENCE OF ODD PERFECT NUMBERS

JOHN VOIGHT

Abstract. In this article, we show how to prove that an odd perfect number with eight distinct prime factors is divisible by 5.

A perfect number N is equal to twice the sum of its divisors: σ(N) = 2N. The theory of perfect numbers when N is even is well known: Euclid proved that if 2p − 1 is prime, then 2p−1(2p − 1) is perfect, and Euler proved that every one is of this type. These numbers have seen a great deal of attention, ranging from very ancient numerology (Saint Augustine considered 6 to be a truly perfect number, since God fashioned the Earth in precisely this many days). They were also very important to the Greeks and to Fermat, whose investigations led him to his little theorem. Today, we have found 38 Mersenne primes (those of the form 2p − 1); the latest, found on June 1, 1999 by Nayan Hajratwala, was part of the Great Internet Mersenne Prime Search (GIMPS) (see http://www.mersenne.org/); it is 26972593 − 1, a number of over two million digits. It is not known if there are inﬁnitely many such primes. Although the collection of increasingly large perfect numbers continues unabated to this day, no odd perfect number has even been found. Worse still, no proof of their non-existence has been given, either. This article reviews the results concerning odd perfect numbers and shows how to prove that an odd perfect number with eight distinct prime factors must be divisible by 5.

1. Known results There are a myriad of known conditions that an odd perfect number N must Qk αi satisfy. Write N = i=1 pi , p1 < ··· < pk, k = ω(N), the number of distinct prime factors of N. • Lower bound: A long computer search by Brent and Cohen (1991) [2] k found that N > 10300. Heath-Brown (1994) [12] proved that N < 44 . • Multiplicative structure: Euler proved that N = $γ m2 where $ ≡ γ ≡ 1 (mod 4). The factor $ is often referred to as the special prime. Touchard [24] proved that if N is an odd perfect number, then N ≡ 1 (mod 9) or N ≡ 9 (mod 36). • Large prime factors: Hagis-Cohen (1998) [10] found that the largest 6 prime factor pk | N satisﬁes pk > 10 . Iannucci (1999) [14], [15] extended the work of Pomerance and Hagis proving that pk−1 > 10000 and pk−2 > 100. These proofs are almost exclusively computational (so can be easily

Date: December 15, 1999; Revised October 29, 2001. 1 2 JOHN VOIGHT

extended with more computing power), and are highly useful in proving other results. • Density: Dickson (1913) [6] proved there are at most finitely many odd perfect numbers with any given number of distinct prime factors. (This also follows from Heath-Brown’s result.) Hornfeck and Wirsing (1957) [13] showed that the number of (odd or even) perfect numbers ≤ x is O(x) for any > 0. • Number of distinct prime factors: E.Z. Chein in his doctoral thesis at Penn State (1979) [3] and Hagis (1980) [8] independently proved that the number of distinct prime factors k = ω(N) ≥ 8. Kishore (1983) [18] and Hagis (1983) [9] proved that if 3 - N then ω(N) ≥ 11. • Small prime factors: Grün(1952) [7] and Perisastri (1958) [20] proved that the smallest factor p1 | N satisfies p1 < (2/3)k + 2. Kishore (1981) 2i−1 [17] proved that pi < 2 (k − i + 1) for 2 ≥ i ≥ 6. αi • Exponents: For pi k N non-special, let αi = 2βi for each i. Steuerwald (1937) proved that βi = 1 for each i is impossible; Kanold (1942) [16] proved that βi = 2 is impossible and that if d = gcd(αi + 1) > 1, then 9, 15, 21, 33 - d. There are other results of this type. While J.J. Sylvester described these numbers as “doubtful or suppositious enti- ties” [23], Descartes was not so pessimistic, giving the spoof example N = 327211213222021 = 198585576189. Since σ(N) = (1 + 3 + 32)(1 + 7 + 72)(1 + 11 + 112)(1 + 13 + 132)(1 + 22021) = = 397171152378 = 2N, N is perfect if we are willing to ignore the fact that 22021 = 19261 is not prime. It is also somewhat surprising that for N = 3472112192(−127), we have (1 + 3 + 32 + 33 + 34)(1 + 7 + 72)(1 + 11 + 112)(1 + 19 + 192)(1 − 127) = = −44035951806 = 2N, which is quite close to saying N is perfect if we are willing to ignore the fact that −127 is negative.

2. Basic Results

Qk αi Let Φd be the dth cyclotomic polynomial. If N = i=1 pi is perfect, then

k k αi+1 k Y αi Y pi − 1 Y Y 2N = σ(N) = σ (pi ) = = Φd(pi). pi − 1 i=1 i=1 i=1 d|αi+1 d>1 This most basic result is the backbone of almost every significant result concerning αi odd perfect numbers, for if pi k N, then every prime factor of Φd(pi) for d | αi + 1 must also divide N. J.J. Sylvester was perhaps the first to effectively use the following function in the analysis of odd perfect numbers: Definition 2.1. We define the abundance of n to be h(n) = σ(n)/n. Proposition 2.2. Suppose p, q are prime. The function h satisfies: ODD PERFECT NUMBERS 3

(a) h is multiplicative and is given by

k Y 1 − 1/pα+1 h(n) = ; 1 − 1/p i=1 α ∞ (b) limα→∞ h(p ) = h(p ) = p/(p − 1); (c) 1 < h(pα) < h(pβ) if 0 < α < β ≤ ∞; (d) h(pα) < h(qβ) if p > q and 0 < α, β ≤ ∞; and (e) N is perfect if and only if h(N) = 2. The proof is an easy exercise. Lemma 2.3 (Nagell [19]). If q is a prime that does not divide n, then

Φn(x) ≡ 0 (mod q) is solvable iﬀ q ≡ 1 (mod n). The solutions of the congruence are exactly the numbers that have order n modulo q, and Φn(x) is divisible by exactly the same power of q as xn − 1. If q | n, let n = qβm, gcd(m, q) = 1. Then

Φn(x) ≡ 0 (mod q) is solvable iﬀ q ≡ 1 (mod m). The solutions are the numbers that have order m 2 modulo q, and Φn(x) is divisible by q and not q whenever n > 2. n Q Writing x − 1 = d|n Φd(x) makes this result clear. Let vq(n) be the maximal vq (n) power of q dividing n; we will also write q k n. Let oq(p) be the order of p (mod q). Then we have: Lemma 2.4. If pα k N, and q ≥ 3 is prime, then

 oq (p) vq(p − 1) + vq(α + 1), oq(p) | (α + 1) and oq(p) 6= 1; α  vq (σ(p )) = vq(α + 1), oq(p) = 1;  0, otherwise. Proof. This result follows from the above and from pα+1 − 1 Y σ(pα) = = Φ (p). p − 1 d d|α+1 d>1 α β If q | σ(p ), then the relevant values of d | (α + 1) occur when d = q oq(p). Suppose oq(p) 6= 1 and oq(p) | α + 1, the ﬁrst case of the lemma. Then the only 1 vq (α+1) values of d that have q | Φd(p) are oq(p), q oq(p), . . . , q oq(p). In the ﬁrst case, it is divisible by a power of q equal to qoq (p) − 1, and in the others it is divisible only by q. Adding these up gives the result. 2 vq (α+1) If oq(p) = 1, then we are interested in d = q, q , . . . , q (Φ1(p) is not in the product) and each of these is divisible only by a single q, again the statement. If oq(p) - α + 1, then p is not a solution to the congruence for any value of d and the product is never divisible by q. There is a central role in this theory by Fermat primes, since in the multiplicative group modulo such a prime, no number can have an odd order. Indeed, an argument like the above proves: 4 JOHN VOIGHT

Lemma 2.5 ([21]). If q is a Fermat prime,  vq(α + 1), p ≡ 1 (mod q); α  vq (σ(p )) = vq(p + 1) + vq(α + 1), p ≡ −1 (mod q), p = $;  0, otherwise. We also have: Proposition 2.6 (Bang, Sylvester [23], Birkhoﬀ-Vandiver [1]). If m is an integer, and m ≥ 2 then Φd(m) is divisible by a prime q with oq(m) = d, except when m = 2 and d = 1 or 6 and when m is a Mersenne number and d = 2. In our case, m will always be an odd prime. Whenever d = 2 we have m ≡ 1 (mod 4) so m is not a Mersenne number, so the exceptional cases will be irrelevant. α Corollary 2.7. If p is a component of N and if d | α + 1, then Φd(p) is divisible by a prime q with oq(p) = d, where q ≡ 1 (mod d). Now consider the following general strategy. Suppose we have a Fermat prime q, and qν k σ(pα) for some component pα of N where ν > 0. Suppose that p is not ν special. Then from Lemma 2.5, we must have vq(α+1) = ν so α+1 = q m, and from the corollary, Φq(p), Φq2 (p),..., Φqν (p) are each divisible by primes, respectively i r1, r2, . . . , rν where the order of p modulo ri is q so that each is distinct, and by i Lagrange ri ≡ 1 (mod q ). If instead we have the special component $γ and qν k σ($γ ), then from the lemma we must have $ ≡ ±1 (mod q). If we let vq(γ + 1) = µ = ν − vq($ + 1), then remembering that 2 | γ+1 (γ is the only odd exponent), we write γ+1 = 2qµm for some m, so N must be divisible by 2µ distinct primes, where in the corollary we take the sums Φq(p), Φ2q(p),..., Φqµ (p), Φ2qµ (p) and corresponding prime divisors 0 0 0 i r1, r1, . . . , rµ, rµ each with distinct orders and like before ri ≡ ri ≡ 1 (mod q ). It is important to note that the preceding conclusions follow just as easily from the fact that p ≡ 1 (mod q) just by inspection of the q-valuation lemmas. We state this as: Proposition 2.8. If p ≡ 1 (mod q) or q is Fermat, then: (a) If p is not special and qν k σ(pα), then N is divisible by distinct primes i r1, . . . , rν where ri ≡ 1 (mod q ). (b) If p = $ is the special prime, qν k σ($γ ) and qµ k (γ + 1), then N is 0 0 0 i divisible by 2µ distinct primes r1, r1, . . . , rµ, rµ where ri ≡ ri ≡ 1 (mod q ). 3. Two Propositions β Proposition 3.1. Suppose that q is a Fermat prime and q | N. Let ωq be the maximum number of factors of N which are 1 (mod q); suppose k of these factors are unknown, and that $ is among the unknown factors. Suppose that at most qν divides σ of the known factors and that β − ν ≥ k(ωq − 1). If ωq ≤ 2, then vq($ + 1) ≥ β − ν; otherwise, vq($ + 1) ≥ β − ν − (k − 1)(ωq − 2) − b(ωq − 2)/2c. In either case, $ is at least equal to the smallest prime $ ≡ −1 (mod qvq ($+1)), ODD PERFECT NUMBERS 5 so in particular $ ≥ 2qvq ($+1) − 1. Proof. By Lemma 2.5, q divides σ(pα) when p is non-special only if p ≡ 1 (mod q), ν and the exponent is exactly vq(α + 1). With q maximally dividing the known factors, we must have qβ−ν dividing σ of the unknown factors. At most we have ωq −1 αi ωq q k σ(pi ) for each such i, since by Proposition 2.8, if q divides such a component then there are at least ωq other distinct prime factors dividing N congruent to 1 (mod q), a contradiction. So qβ−ν−(k−1)(ωq −1) | σ($γ ), the special prime. γ Now if $ ≡ 1 (mod q), then vq($ ) = vq(γ + 1) = µ, and we know 2µ ≤ ωq − 1 again counting primes 1 (mod q). Weakening this slightly, we can say µ < ωq − 1, and adding these up, we have β−ν ≥ k(ωq −1) > (k−1)(ωq −1)+µ, a contradiction since we then have factors of q unaccounted for. So $ ≡ −1 (mod q), and we can strengthen the above analysis to get at most ωq −2 α q k σ(p ) (since the input guess of ωq was one too high). If ωq ≤ 2, this gives us the fact that we already knew: if q | σ(N/$γ ), and $ ≡ −1 (mod q), then there are at least two primes congruent to 1 (mod q), and if this cannot hold, then qβ−ν k σ($γ ). Otherwise, µ = vq(γ + 1) < (ωq − 1)/2, so since µ is an integer, µ ≤ b(ωq − 2)/2c and at most qb(ωq −2)/2c | (γ + 1), so

vq($ + 1) = β − ν − (k − 1)(ωq − 2) − b(ωq − 2)/2c, with appropriate replacements when ωq < 2. Remembering that $ is odd we must have $ + 1 ≥ 2qvq ($+1) as claimed. As a direct consequence, we have the special case: Corollary 3.2. Suppose each of the conditions in Proposition 3.1 is satisﬁed, and we have qν k σ(N/$γ )—i.e., k = 1. Then $ ≥ 2qβ−ν−b(ωq −2)/2c − 1. In the case that the Fermat prime q = 3, we can also get a very large value for q7: Corollary 3.3. Suppose in the proposition we have either q = 3 or qβ k $ + 1. Let β P be the largest prime factor dividing σ(q ). Then pk−1 ≥ min($, P ). Proof. Suppose q = 3. Then since $ ≡ −1 (mod 3), but $ ≡ 1 (mod 4), we have 3 $ = = −1, $ 3 using the Legendre symbol; but then $ | σ(3α) says 3α+1 ≡ 1 (mod $) and since α + 1 is odd, we can multiply both sides by 3 and get that 3 is a square, a contradiction. Therefore, P 6= $, and the result follows. The second case follows from $ + 1 ≥ 2qβ > σ(qβ) so that P 6= $. This is a useful result because: 6 JOHN VOIGHT

Lemma 3.4 (Hagis and McDaniel [11]). Let α + 1 be an odd prime, and let P be α the largest prime factor of Φα+1(q) = σ(q ). Then: (a) If q = 3, then P ≥ 100129 except for α ∈ {2, 4, 6, 10, 16}; (b) If q = 5, then P ≥ 100129 except for α ∈ {2, 4, 6}; and (c) If q = 17, then P ≥ 88741 except for α = 2. I improved this in the case q = 3: Lemma 3.5. If in the conditions of Lemma 3.4 we have q = 3, then in fact P ≥ 363889. Proof. For each prime p = α + 1, it suﬃces to look at primes q ≡ 1 (mod p) such that q < 363889/2 (by Corollary 2.7) and such that 3p ≡ 1 (mod qβ) for some β p−1 β > 0 (by Lemma 2.5). If the product of the q is equal to σ(3 ) = Φp(3), then this factors completely under 363889; otherwise, it has a larger prime factor. Obtaining this result then reduces to a number of power computations in modular arithmetic. In the case that the special prime is one of the known factors, we can argue as follows:

β Proposition 3.6. Suppose that q is a Fermat prime and q | N. Let ωq be the maximum number of prime factors of N which are 1 (mod q); suppose that k factors of these are unknown, and that $ is among the known factors. Suppose that at most qν divides σ of the known factors. Then at most qk(ωq −1)+ν k N. Proof. Just like in Proposition 3.1, we have qβ−ν dividing the known factors, and ωq −1 αi k(ωq −1) at most we have q k σ(pi ) for an unknown factor pi. Hence at most q divides the unknown components, and we have the result since β−ν = k(ωq −1). 4. Conclusion From these propositions, we can extend the work of Hagis and Chein, and prove: Theorem 4.1. If N is perfect and ω(N) = 8, then 5 | N. The proof is an extended computation, the results of which are available from the author. As an illustration, we include a portion of the proof. We know from Kishore [18] and Hagis [9] that 3 | N, from Hagis-Cohen [10] that 6 p8 > 10 , and from Iannucci [14], [15] that p6 > 100, p7 > 10000. First:

Lemma 4.2. Either p2 = 5 and p3 ≤ 47, p2 = 7 and p3 ≤ 19, or p2 = 11 and p3 = 13. Proof. Recall the properties of the abundance function h listed in Proposition 2.2. Suppose 5, 7, 11 - N. Then h(N) < h(3∞13∞17∞19∞23∞101∞10007∞1000003∞) < 2, a contradiction. Hence p2 ≤ 11. In each of these cases, h(3∞5∞53∞59∞61∞101∞10007∞1000003∞) < 2 h(3∞7∞23∞29∞31∞101∞10007∞1000003∞) < 2 h(3∞11∞17∞19∞23∞101∞10007∞1000003∞) < 2 gives an upper bound on p3. ODD PERFECT NUMBERS 7

Lemma 4.3. If p2 = 11, then p5 = 19.

Proof. We have p3 = 13, and h(3∞11∞13∞19∞23∞101∞10007∞1000003∞) < 2 implies p4 = 17 and h(3∞11∞13∞17∞29∞101∞10007∞1000003∞) < 2 implies p5 = 19 or 23. We have two cases. First suppose p5 = 23. From h(3∞11∞13∞17∞23∞149∞10007∞1000003∞) < 2 we have p6 ≤ 139. Now h(3411∞13∞17∞23∞101∞10007∞1000003∞) < 2

6 6 so 3 | N. But Φ7(3) = σ(3 ) = 1093, and h(3611∞13∞17∞23∞1093∞10007∞1000003∞) < 2

2 implies 17 | N, and since Φ3(17) = 307, h(3∞11∞13∞17223∞307∞10007∞1000003∞) < 2 implies 174 | N. Now

h(N) > h(381141341742321312) > 2 implies p6 ≥ 137. Suppose 137 | N. Suppose $ = 137. We would like to use Proposition 3.6. We have k = 2 unknown factors. Also, at most 3 singly divides σ of the known factors 2 because σ(13 ) = Φ3(13) has an impossible from the above, and 11, 17, 23, 137 ≡ 2 (mod 3), 3 k (137 + 1): recall from the Fermat valuation lemma that 3 | σ(pα) iﬀ p ≡ 1 (mod 3) and 3 | α + 1. Thus ν = 1, and ω3 = 3 so by the proposition we 2(3−1)+1 5 have at most 3 = 3 k N, a contradiction. Now since 7 | Φ3(23), and h(381141341742341372) > 2.

Thus p6 = 139. We can now rule out $ = 17 since with ω3 = 4, ν = 3 2 9 (499 | Φ3(139) and 3 k Φ2(17)), we conclude 3 k N, and 1093 = Φ7(3), 757 | Φ9(3) cannot occur so we have a contradiction. Now since 139 6≡ 1 (mod 17), we know from Corollary 3.3 that p7 ≥ 88741, and h(3∞11∞13∞17∞23∞139∞88741∞1000003∞) < 2, a contradiction.

One can continue in this way to prove the theorem. 8 JOHN VOIGHT

References [1] George D. Birkhoff and H.S. Vandiver, On the integral divisors of an − bn, Ann. of Math. 5 (1904), 173–180. [2] R.P. Brent, G.L. Cohen, H.J.J. te Riele, Improved techniques for lower bounds for odd perfect numbers, Math. Comp. 57 (1991), 857–868. [3] Joseph E.Z. Chein, An odd perfect number has at least 8 prime factors, Ph.D. thesis, Penn- sylvania State University, 1979. [4] G.L. Cohen, On odd perfect numbers, Fibonacci Q. 16 (1978), 523–527. [5] L.E. Dickson, Notes on the theory of numbers, Amer. Math. Monthly 18 (1911), 109. [6] L.E. Dickson, Finiteness of the odd perfect and primitive abundant numbers with n distinct prime factors, Amer. J. Math. 35 (1913), 413–422. [7] Otto Grün, Uber¨ ungerade vollkommene Zahlen, Math. Zietschr. 55 (1952), 353–354. [8] Peter Hagis, Jr., Outline of a proof that every odd perfect number has at least eight prime factors, Mat. Comp. 35 (1980), 1027–1032. [9] Peter Hagis, Jr., Sketch of a proof that an odd perfect number relatively prime to 3 has at least eleven factors, Math. Comp. 40 (1983), 399-404. [10] Peter Hagis, Jr. and Graeme L. Cohen, Every odd perfect number has a prime factor which exceeds 106, Math. Comp. 67 (1998), 1323–1330. [11] Peter Hagis, Jr. and Wayne L. McDaniel, On the largest prime divisor of an odd perfect number. II, Math. Comp. 29 (1975), 922–924. [12] D.R. Heath-Brown, Odd perfect numbers, Math. Proc. Cambridge Phil. Soc. 115 (1994), 191–196. [13] Bernhard Hornfeck and Eduard Wirsing, Uber¨ die Häufigkeit vollkommener Zahlen, Math. Ann. 133 (1957), 431–438. [14] Douglas E. Iannucci, The second largest prime divisor of an odd perfect number exceeds ten thousand, Math. Comp. 68 (1999), 1749–1760. [15] Douglas E. Iannucci, The third largest prime divisor of an odd perfect number exceeds one hundred, Math. Comp. 69 (2000), 867–879. [16] Hans-Joachim Kanold, Verschärfungeiner notwendigen Bedingung fürdie Existenz einer ungeraden vollkommenen Zahl, J. Reine Angew Math. 184 (1942), 116–123. [17] Masao Kishore, On odd perfect, quasiperfect, and odd almost perfect numbers, Math. Comp. 36 (1981), 583–586. [18] Masao Kishore, Odd perfect numbers not divisible by 3. II, Math. Comp. 40 (1983), 405–411. [19] Trygve Nagell, Introduction to number theory, New York: Wiley & Sons, 1951, 164–168. [20] M. Perisastri, A note on odd perfect numbers, Math. Student 26 (1958), 179–181. [21] Carl Pomerance, Odd perfect numbers are divisible by at least seven distinct primes, Acta. Arith. 25 (1974), 265–300. [22] D. Suryanarayana and Peter Hagis, Jr., A theorem concerning odd perfect numbers, Fi- bonacci Q. 8 (1970), 337–346, 374. [23] J.J. Sylvester, On the divisors of the sum of a geometrical series whose first term is unity and a common ratio any positive or negative integer, Nature 37 (1888), 417–418. [24] Jacques Touchard, On prime numbers and perfect numbers, Scripta Math. 19 (1953), 35–39.

Department of Mathematics, University of California, Berkeley E-mail address: [email protected]